A) 10 years and 20 years, respectively
B) 5 years and 20 years, respectively
C) 20 years and 10 years, respectively
D) 20 years and 5 years, respectively
Correct Answer: B
Solution :
| [b] For sample 1 we can write, |
| \[{{N}_{1}}={{N}_{o1}}{{e}^{-{{\lambda }_{1}}t}}\] |
| \[{{N}_{2}}={{N}_{o2}}{{e}^{-{{\lambda }_{2}}t}}\] |
| Given that at a particular time t\[{{N}_{1}}=2{{N}_{2}}\] |
| Also given that |
| For sample 1, |
| \[{{A}_{1}}=-\frac{d{{N}_{1}}}{dt}={{N}_{o1}}{{\lambda }_{1}}{{e}^{-{{\lambda }_{1}}t}}=5\mu {{C}_{i}}\] |
| \[{{A}_{2}}=-\frac{d{{N}_{2}}}{dt}={{N}_{o2}}{{\lambda }_{2}}{{e}^{-{{\lambda }_{2}}t}}=10\mu {{C}_{i}}\] |
| Solving above equations we get, |
| \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{1}{4}\] |
| \[\frac{half\,life\,{{e}_{1}}}{half\,life\,{{e}_{2}}}=\frac{\ln 2}{{{\lambda }_{1}}}\times \frac{{{\lambda }_{2}}}{\ln 2}=4\] |
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